Mario rolls a coin up a slope at 3 m/s North. It travels 8.3 m, comes to a stop and rolls back down.

The body is subjected to a force of 0,4 N m with a shoulder of 5 cm. What is the magnitude of this force?

igor_vitrenko [27]

Answer:

8N

Explanation:

the body is subjected to pressure equals to force force inverse to the area of that body given that it is in motion

5 0

2 years ago

Tom kicks a soccer ball on a flat, level field giving it an initial speed of 20 m/s at an angle of 35 degrees above the horizont

SVEN [57.7K]

Answer:

(a) 2.34 s

(b) 6.71 m

(c) 38.35 m

(d) 20 m/s

Explanation:

u = 20 m/s, theta = 35 degree

(a) The formula for the time of flight is given by

$T = \frac{2 u Sin\theta }{g}$

$T = \frac{2 \times 20 \times Sin35 }{9.8}$

T = 2.34 second

(b) The formula for the maximum height is given by

$H = \frac{u^{2} \times Sin^{2}\theta }{2g}$

$H = \frac{20^{2} \times Sin^{2}35 }{2 \times 9.8}$

H = 6.71 m

(c) The formula for the range is given by

$R = \frac{u^{2} \times Sin 2\theta }{g}$

$R = \frac{20^{2} \times Sin 2 \times 35}{9.8}$

R = 38.35 m

(d) It hits with the same speed at the initial speed.

8 0

3 years ago

*WILL MARK BRAINLIEST*

Arte-miy333 [17]

Answer:

1.52

Explanation:

5 0

3 years ago

Read 2 more answers

Can a goalkeeper at his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will

zmey [24]

The goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

Explanation:

Consider the vertical motion of ball,

We have equation of motion v = u + at

Initial velocity, u = u sin θ

Final velocity, v = 0 m/s

Acceleration = -g

Substituting

v = u + at

0 = u sin θ - g t

$t=\frac{usin\theta }{g}$

This is the time of flight.

Consider the horizontal motion of ball,

Initial velocity, u = u cos θ

Acceleration, a =0 m/s²

Time, $t=\frac{usin\theta }{g}$

Substituting

s = ut + 0.5 at²

$s=ucos\theta \times \frac{usin\theta }{g}+0.5\times 0\times (\frac{usin\theta }{g})^2\\\\s=\frac{u^2sin\theta cos\theta}{g}\\\\s=\frac{u^2sin2\theta}{2g}$

This is the range.

In this problem

u = 30 m/s

g = 9.81 m/s²

θ = 45° - For maximum range

Substituting

$s=\frac{30^2\times sin(2\times 45)}{2\times 9.81}=45.87m$

Maximum horizontal distance traveled by ball without touching ground is 45.87 m, which is less than 95 m.

So the goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

6 0

3 years ago

A sprinter accelerates from rest to 10.0 m/s in 1.35 s l. What is her acceleration?

ale4655 [162]

Answer:

dsfsdfgfdsggfgsgsfsfgfsgffsfgfgfgf

Explanation:

sgdfggsfdsgfgsgsmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

8 0

3 years ago